Q.Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 °C to ice at –10.0 °C. $\Delta_{fus}H = 6.03$ kJ mol$^{-1}$ at 0 °C. $C_p[H_2O(l)] = 75.3$ J mol$^{-1}$ K$^{-1}$; $C_p[H_2O(s)] = 36.8$ J mol$^{-1}$ K$^{-1}$.

Concept: Enthalpy change for a multi-step process (cooling liquid → freezing → cooling solid).

1. Cool liquid water from $10.0\,^\circ\text{C}$ to $0\,^\circ\text{C}$:

   $\Delta H_1 = n C_p(l) \Delta T = (1.0)(75.3)(-10.0) = -753\ \text{J} = -0.753\ \text{kJ}$

2. Freeze water at $0\,^\circ\text{C}$ (exothermic):

   $\Delta H_2 = -n \Delta_{fus}H = -(1.0)(6.03) = -6.03\ \text{kJ}$

3. Cool ice from $0\,^\circ\text{C}$ to $-10.0\,^\circ\text{C}$: …

The problem asks for the enthalpy change when 1.0 mol of water freezes — but the water starts at 10 °C and the ice ends at –10 °C. You cannot simply use the enthalpy of fusion at 0 °C, because the temperature changes before and after freezing also carry enthalpy changes. The key insight: enthalpy is a state function, so we can design any convenient path that goes from the initial state (liquid, 10 °C) to the final state (solid, –10 °C) and sum the enthalpy changes along that path. The actual path doesn’t matter — only the start and end states do.

We choose a three‑step path:

1. Cool the liquid water from 10 °C to 0 °C.
2. Freeze the water at 0 °C (the reverse of fusion, so the sign flips).
3. Cool the ice from 0 °C to –10 °C.

Each step uses the heat capacity or the enthalpy of phase change. Let’s work through them.

Step 1: Cooling liquid water from 10 °C to 0 °C

The enthalpy change for a temperature change at constant pressure is $\Delta H = n C_p \Delta T$. Here $n = 1.0$ mol, $C_p(\text{liquid}) = 75.3$ J mol⁻¹ K⁻¹, and $\Delta T = T_f - T_i = 0 - 10 = -10$ K.

The negative sign makes sense: cooling releases heat, so enthalpy decreases.

Step 2: Freezing at 0 °C

The enthalpy of fusion (melting) is $\Delta_{fus}H = +6.03$ kJ mol⁻¹. Freezing is the reverse process, so the enthalpy change for freezing is the negative of that:

Again, negative because freezing is exothermic.

Step 3: Cooling ice from 0 °C to –10 °C

Now we use the heat capacity of solid water: $C_p(\text{solid}) = 36.8$ J mol⁻¹ K⁻¹. The temperature change is $\Delta T = -10 - 0 = -10$ K.

$\Delta H_3 = (1.0)(36.8)(-10) = -368\ \text{J} = -0.368\ \text{kJ}$ …